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Background

Suppose that we are using a simplified spherical model of the Earth's surface with latitude $u \in (-\frac {\pi} 2, \frac {\pi} 2)$ and longitude $v \in (-\pi, \pi)$; then (if the radius is taken to be $1$), the surface area element is given by $\mathrm{d}A = \cos u\ \mathrm{d}u\ \mathrm{d}v$. Restricting attention to the hemisphere, $H$, where $u, v \in (-\frac {\pi} 2, \frac {\pi} 2)$, a simple map projection from $H$ can be obtained by just taking the $x$ and $y$ coordinates via $x = \cos u \sin v$ and $y = \sin u$, which is a smooth one-to-one transformation on $H$. Now, picking a point with coordinates $(U, V)$ on $H$ uniformly according to surface area, the joint density of $U$ and $V$ is $$f_{U, V}(u, v) = \frac 1 {2\pi} \cos u, \quad \lvert u \rvert, \lvert v \rvert < \frac {\pi} 2.$$

Question

Find the joint density of $X$ and $Y$, where $(X, Y)$ is the image of the random point $(U, V)$ under the map projection defined above.

My working

Some preliminary results:

$$\begin{aligned} J & = \begin{vmatrix} \frac {dx} {du} & \frac {dx} {dv} \\ \frac {dy} {du} & \frac {dy} {dv} \end{vmatrix} \\[1 mm] & = \begin{vmatrix} -\sin u \sin v & \cos u \cos v \\ \cos u & 0 \end{vmatrix} \end{aligned}$$

Thus,

$$\begin{aligned} \\[1 mm] f_{X, Y}(x, y) & = \frac 1 {2\pi} \cos u\ \lvert J \rvert \\[1 mm] & = \frac 1 {2\pi} \cos^3 u \cos v \\[1 mm] & = \frac 1 {2\pi} x(1 - y^2) \cot v \\[1 mm] & = \frac 1 {2\pi} (1 - y^2) \sqrt {1 - y^2 - x^2} \end{aligned}$$

I am stuck here. Firstly, may I know if my expression for $f_{X, Y}(x, y)$ is correct? Secondly, if it is correct, how can I obtain the support for $f_{X, Y}(x, y)$?

Ethan Mark
  • 2,167

1 Answers1

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We have$$f_{X,Y}(x,y)=f_{U,V}(u(x,y),v(x,y))\color{red}{/|}J\color{red}|$$where $J=\frac{\partial(x,y)}{\partial(u,v)}$. I have typed in red the parts where you have faltered. You have to divide by the absolute value of the Jacobian if you calculate it as $\frac{\partial(x,y)}{\partial(u,v)}$.

Thus $$f_{X,Y}(x,y)=\frac{\cos(u(x,y))}{2\pi\cos^2(u(x,y))|\cos(v(x,y))|}$$and since $v\in(-\pi/2,\pi/2),\cos v>0$, we get$$f_{X,Y}(x,y)=\frac1{2\pi\cos(u(x,y))\cos(v(x,y))}=\frac1{2\pi[\cos u\cos v](x,y)}$$and you obtained $\cos v= \frac {\sqrt {\cos^2 u - x^2}} {\cos u}$ or $\cos v \cos u= \sqrt{ \cos^2 u - x^2} =\sqrt{1-y^2-x^2}$, we get$$f_{X,Y}(x,y)=\frac1{2\pi\sqrt{1-x^2-y^2}}$$over the circle $x^2+y^2<1$ and zero otherwise. The support was obtained using the constraints on $u,v$ in $f_{U,V}$: $$\begin{align*} |u|<\pi/2&\implies y=\sin u\in(-1,1)\\ |v|<\pi/2&\implies \frac x{\cos u}=\frac x{\sqrt{1-y^2}}=\sin v\in(-1,1)\\ &\implies\left|\frac x{\sqrt{1-y^2}}\right|<1\\ &\implies x^2<1-y^2 \end{align*}$$

Shubham Johri
  • 17,659
  • Just remember one property of the Jacobian, that $\frac{\partial(x,y)}{\partial(u,v)}=\left[\frac{\partial(u,v)}{\partial(x,y)}\right]^{-1}$. So you either divide by $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|$ or alternatively multiply by $\left|\frac{\partial(u,v)}{\partial(x,y)}\right|$, depending on whether it is easy to obtain $u,v$ as functions of $x,y$ for finding $u_x,u_y,v_x,v_y$. In this example it was not, $x_u,x_v,y_u,y_v$ were readily calculable, so we divide. Ultimately the formula is$$f_{X,Y}(x,y)=f_{U,V}(u(x,y),v(x,y))\left|\frac{\partial(u,v)}{\partial(x,y)}\right|$$ – Shubham Johri May 20 '21 at 12:45
  • A good way to remember is that $f_{U,V}$ will get multiplied with $\left|\frac{\partial(u,v)}{\partial(x,y)}\right|$, i.e. with the $u,v$ in the numerator of the Jacobian notation. – Shubham Johri May 20 '21 at 12:46
  • I see. So basically I can remember it in the following way - that the numerator of the Jacobian must always follow the variables of the original joint density? – Ethan Mark May 20 '21 at 12:49
  • Yes, that is how at-least I remember it Ethan – Shubham Johri May 20 '21 at 12:50
  • Right. Thank you! One last thing... in determining the support for $f_{X, Y}(x, y)$, do we have to do it your way, or can we just observe directly that $x^2 + y^2 <$ because of the square root in the denominator? – Ethan Mark May 20 '21 at 13:12
  • @EthanMark Here one can predict that the support is some subset of the region $x^2+y^2<1$, and luckily the support is the entire region. But that is not the general case. You may not always be able to tell the support from the expression of the joint distribution, so the general way is to deduce it from the constraints. – Shubham Johri May 20 '21 at 13:53
  • Hi! I have a follow-up query to this problem, which I have posted in another question here. In your free time, would you be able to take a look at it? – Ethan Mark May 22 '21 at 08:25
  • Sure, I am taking a look. – Shubham Johri May 22 '21 at 08:25